java 如何修复 Kotlin 中的重载分辨率歧义（无 lambda）？

How to fix Overload Resolution Ambiguity in Kotlin (no lambda)?

提问by Berry

I am having Overload Resolution Ambiguity error in this line:

我在这一行遇到了重载解析歧义错误：

departureHourChoice!!.selectionModel.select(currentHourIndex)

departureHourChoice!!.selectionModel.select(currentHourIndex)

For Reference:

以供参考：

• departureHourChoiceis a ChoiceBox<Int>, which is from java.scene.control

• currentHourIndexis an Int

• The Overload Resolution Ambiguity happens in the .select()method; It is overloaded and can accept two kinds of parameters: (T obj)or (int index).

• The .select()method allows for an item in a ChoiceBoxto be selected, and you can determine which one can be selected by referencing to that item or to it's index. In this case, I want it to be selected by Index (int).

• Here is a photo of the error

• departureHourChoice是 a ChoiceBox<Int>，它来自java.scene.control

• currentHourIndex是一个 Int

• 重载解析歧义发生在.select()方法中；它是重载的，可以接受两种参数：(T obj)或(int index)。

• 该.select()方法允许ChoiceBox选择a 中的项目，您可以通过引用该项目或其索引来确定可以选择哪个项目。在这种情况下，我希望它由 Index ( int)选择。

• 这是错误的照片

How would one resolve the Overload Resolution Ambiguity?

如何解决重载解析歧义？

采纳答案by miensol

It seems that you are hit by this bugas a workaround you can :

似乎您被此错误击中，作为一种解决方法，您可以：

• box the currentHourIndex:

  lateinit var departureHourChoice: ChoiceBox<Int>
  ...
  val currentHourIndex = 1
  departureHourChoice.selectionModel.select(currentHourIndex as Int?)
  

• or change declaration of ChoiceBoxto use java.lang.Integerinstead of Kotlin's Int:

  lateinit var departureHourChoice: ChoiceBox<java.lang.Integer>
  ...
  val currentHourIndex = 1
  departureHourChoice.selectionModel.select(currentHourIndex)
  

• 框currentHourIndex：

  lateinit var departureHourChoice: ChoiceBox<Int>
  ...
  val currentHourIndex = 1
  departureHourChoice.selectionModel.select(currentHourIndex as Int?)
  

• 或更改声明ChoiceBox为 usejava.lang.Integer而不是 Kotlin 的Int：

  lateinit var departureHourChoice: ChoiceBox<java.lang.Integer>
  ...
  val currentHourIndex = 1
  departureHourChoice.selectionModel.select(currentHourIndex)
  

Further reading:

进一步阅读：

• Why is Integer parameter of Java method mapped to Int and not platform type?
• Kotlin: What can I do when a Java library has an overload of both primitive and boxed type?

• 为什么 Java 方法的 Integer 参数映射到 Int 而不是平台类型？
• Kotlin：当 Java 库同时具有原始类型和装箱类型的重载时，我该怎么办？

回答by Rowan Gontier

The solution for me in similar situation is to define in your import for example: import kotlin.math.sqrt as kotsqrt

在类似情况下，我的解决方案是在您的导入中定义，例如：import kotlin.math.sqrt as kotsqrt

then use as: val a = kotsqrt(2.3)

然后用作：val a = kotsqrt(2.3)

回答by voddan

Try casting to Int:

尝试投射到Int：

departureHourChoice!!.selectionModel.select(currentHourIndex as Int)